Is Every Group of Order p^2 Abelian?

[titaniumx3]

Any element x of order p sits inside a Sylow p-subgroup P. So C(x) = N(x) >= N(P). Now think about how the third Sylow theorem (as stated in wikipedia) helps us.

Apart from what I've written in my previous reply I'm not entirely sure where to go with this.

If [G: C(x)] = np (where np is the number of Sylow p-subgroups in G) then could you argue that [G:C(x)] is coprime to P since np divides m, where m = |G|/(p^n) (according the 3rd part of Sylows Theorem) and we know that p does not divide m.

For the case where [G: C(x)] < np I'm not sure what to argue.

 

[JasonRox]

However, 47 is a prime, and we know that

$$\frac{|G|}{|C_G(a)|}$$ cannot be a prime, thus case 3 cannot happen, and so in all cases, we get that G is simple

Why can't the sum of all of them be prime?

 

[titaniumx3]

Why can't the sum of all of them be prime?

Yeah, the summation seems to be missing, I didn't see that the first time round.

 

[morphism]

Sorry, what I'd previously told you isn't correct. "N(x) >= N(P)" is not necessarily true at all.

 

[Kreizhn]

Then think about it like this. What are the possible order of the conjugacy classes? Then what numbers must we use to sum up to 47, especially since [tex][G][/tex] must divide $$|G|$$. Furthermore, we know that if [tex] |G| \text{ does not divide } [G]! [/tex] for any $$H < G$$ then G is not simple.

As a matter of fact, I think there may be a result that stipulates by this same reasoning that if the order of the group is $$|G| = mp, \; m\in\mathbb{N}, \; p$$ a prime, then G is not simple. I'm not sure about this one though, I'll have to do a bit more looking.

 

[JasonRox]

As a matter of fact, I think there may be a result that stipulates by this same reasoning that if the order of the group is $$|G| = mp, \; m\in\mathbb{N}, \; p$$ a prime, then G is not simple. I'm not sure about this one though, I'll have to do a bit more looking.

Well, look at the alternating group A_5. It's simple but the order of A_5 is 60 which is 30*2 where 2 is the p (prime) and 30 is the m (the natural number).

 

[morphism]

Then think about it like this. What are the possible order of the conjugacy classes? Then what numbers must we use to sum up to 47, especially since [tex][G][/tex] must divide $$|G|$$. Furthermore, we know that if [tex] |G| \text{ does not divide } [G]! [/tex] for any $$H < G$$ then G is not simple.

Kreizhn, this problem was solved twice already: in posts #20 and #21.

 

[Dick]

Kreizhn, this problem was solved twice already: in posts #20 and #21.

Apparently, it's Death Proof.

 

[morphism]

Well, look at the alternating group A_5. It's simple but the order of A_5 is 60 which is 30*2 where 2 is the p (prime) and 30 is the m (the natural number).

Maybe Kreizhn meant |G|=mp where gcd(m,p)=1 (and m>1, because it's really trivial otherwise!).

But even in this case it's false: there is a simple group of order 168=3*56.

 

[titaniumx3]

I still can't figure out the last question I posted, which was:

Let G be a finite group. For every prime divisor p of |G| show that there exists an element x in G of order p (i.e. o(x) = p) such that [G: C_G(x)] and p are coprime.

(note: C_G(x) denotes the centraliser of x in G.)


I'm pretty sure you have to use Sylow's theorem but I can only get as far as showing that there exists an element x of order P, that lies in Sylow p-subgroup.

 

[matt grime]

A sylow P subgroup has order a power of p. Take any element in it. What is its order? If it's not p can you think of a way to get an element that does have order p from it?

 

[titaniumx3]

A sylow P subgroup has order a power of p. Take any element in it. What is its order? If it's not p can you think of a way to get an element that does have order p from it?

If the element is not of order p then it is either the identity element or an element that that is of order that divides (p^n) (the order of the Sylow P subgroup). So for some element of order (p^i) raise it to some power j such that (p^i)^j) = p, which is contained in the Sylow p-subgroup.

EDIT: that is false.

 

[matt grime]

Erm, x has order p^i, so if you raise it to the power p^i, you get e, the identity element. Moreover,

x^(p^i +1) is not what you say it is - I think you omitted some brackets and I presume you meant p^{i+1}. Well, x^p^{i+1} is certainly not equal to p, anyway - one is a group element (the identity) and the other is a prime number.

 

[titaniumx3]

Yeah I completely lost it there, sorry.

Anyway, can't you just apply the following corollary to the Sylow p-subgroup:

Given a finite group G and a prime number p dividing the order of G, then there exists an element of order p in G .

Hence there exists a element of order P in the Sylow p-subgroup since p divides it's order.

 

[matt grime]

Of course you can. But since you probably proved that there is such an element by showing that it is true for Sylow subgroups, then it is quite possibly a circular argument.

It is really a simple exercise to prove it for p groups, so you should.

If the order of x is p^r, write down an element of order p - i,.e. a non-identity element which when raised to the p'th power is the identity.

 

[Kummer]

I still can't figure out the last question I posted, which was:

Let G be a finite group. For every prime divisor p of |G| show that there exists an element x in G of order p (i.e. o(x) = p) such that [G: C_G(x)] and p are coprime.
.

(Accept strong induction.)
You can use the conjugacy class equation,
$$|G| = |Z(G)|+\sum [G:C(x)]$$
Where $$x$$ is the sum of all non-trivial conjugacy classes.
Now $$p$$ divides $$|G|$$. We have two cases: Case#1 is that there is an index $$[G:C(x)]$$ is not divisible by $$p$$. Case#2 is that all indexes $$[G:C(x)]$$ are divisible by $$p$$. If Case#1 then $$p$$ does not divide $$[G:C(x)]$$ where $$x$$ is some element forming a non-trivial conjugacy class. Since $$[G:C(x)]|C(x)| = |G|$$ it means $$p$$ must divide $$C(x)$$. If Case#2 then since the sum is divisible by $$p$$ it implies $$Z(G)$$ must be divisible by $$p$$ if $$G$$ is non-abelian i.e. $$Z(G)<G$$ then by strong induction there is an element of order $$p$$ in $$Z(G)$$. So the problem reduces to working with abelian $$G$$. Now let $$a$$ (different from identity) be an element and form the subgroup $$H=\left< a \right>$$ if $$H=G$$ then the group is cyclic and the proof is complete. Otherwise $$H<G$$. Now if $$p$$ divides $$|H|$$ then proof is complete by strong induction so it is safe to say $$H$$ is not divisible by $$p$$. Now form the factor group $$G/H$$ (note $$H$$ is normal since $$G$$ is abelian) it is divisible by $$p$$ since $$|G/H||H|=|G|$$ and $$p$$ does not divide $$H$$. So $$G/H$$ must contain an element $$b$$ which has order $$p$$ by strong induction, i.e. $$(bH)^p \in H$$. We will not claim that $$b^m$$ has order $$p$$ (where $$m=|H|$$). First $$(b^m)^p = (b^p)^m = e$$ since $$b^p\in H$$. Second we show this is the smallest exponent because if it was not then $$b^m = e$$ but then $$(bH)^m = H \in G/H$$ so $$p$$ would be a divisor of $$m$$ a contradiction. So $$b^m$$ has order $$p$$.